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____ [38]
3 years ago
6

All of the following are equal to Avogadro's number EXCEPT ____. a. the number of atoms of bromine in 1 mol Br2 b. the number of

atoms of gold in 1 mol Au c. the number of molecules of nitrogen in 1 mol N2 d. the number of molecules of carbon monoxide in 1 mol CO
Chemistry
1 answer:
Colt1911 [192]3 years ago
8 0

All the following are equal to Avogadro's number EXCEPT a. the number of atoms of bromine in 1 mol Br₂.

1 mol Br₂ contains Avogadro’s number of molecules of Br₂.

However, each molecule contains two atoms of Br, so there are

<em>2 × Avogadro’s number of Br atoms </em>in 1 mol Br₂.

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NEED HELP ASAP! Newton's Law of Gravity says that any the effect of gravity between two objects depends on what two factors?
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Answer:

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Explanation:

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3 years ago
Chem timed exam, help plz
Alchen [17]

Answer:

3.864 g of C₃H₆

Explanation:

The balanced equation for the reaction is given below:

2C₃H₆ + 9O₂ —> 6CO₂ + 6H₂O

From the balanced equation above,

2 moles of C₃H₆ reacted to produce 6 moles of CO₂

Next, we shall determine the number of mole of C₃H₆ that reacted to produce 0.276 mole of CO₂. This can be obtained as follow:

From the balanced equation above,

2 moles of C₃H₆ reacted to produce 6 moles of CO₂.

Therefore, Xmol of C₃H₆ will react to produce 0.276 moles of CO₂ i.e

Xmol of C₃H₆ = (2 × 0.276)/6

Xmol of C₃H₆ = 0.092 mole

Finally, we shall determine the mass of 0.092 mole of C₃H₆. This can be obtained as follow:

Mole of C₃H₆ = 0.092 mole

Molar mass of C₃H₆ = (12×3) + (6×1)

= 36 + 6

= 42 g/mol

Mass of C₃H₆ =?

Mass = mole × molar mass

Mass of C₃H₆ = 0.092 × 42

Mass of C₃H₆ = 3.864 g

Therefore, 3.864 g of C₃H₆ is needed for the reaction.

5 0
3 years ago
Force can be added together
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7 0
3 years ago
The half-life of tritium, or hydrogen-3, is 12.32 years. After about 24.6 years, how much of a sample of tritium will remain unc
scoundrel [369]

Answer: The correct option is B.

Explanation: This is an example of radioactive decay and all the radioactive decay processes follow First order of kinetics.

Expression for the half life of first order kinetics is:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=12.32years

Putting in above equation, we get:

12.32=\frac{0.693}{k}\\k=0.05625year^{-1}

Expression to calculate the amount of sample which is unchanged is:

N=N_oe^{-kt}

where,

N = Amount left after time t

N_o = Initial amount

k = Rate constant

t = time period

Putting value of k = 0.05625 and t = 24.6 in above equation, we get:

N=N_oe^{-0.05625\times 24.6}

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The above fraction is the amount of sample unchanged and that is equal to \frac{1}{4}

Hence, the correct option is B.

8 0
3 years ago
Read 2 more answers
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